If a set is separable its completion is separable

190 Views Asked by At

$\tilde{M}$ is the completion of M.

This question has ben done before, but with no answer. My idea is to first: show that $S$,$T$ sets (which compose the separation of M) are actually the preimage of the $S'$,$T'$ sets (which compose a separation in $\tilde{M}$), I've got an isometry to do this but I can't fill the holes:

1)Show that $S'$ and $T'$ are such that $S'\cap T'=\emptyset$.

2)show that the union of $S'$ and $T'$ are equal to $\tilde{M}$

3)Show isometry moves open sets into open sets ( I'm working on this one)

which would complete th demostration

Am i on the right direction? or am I missing a theorem...?